Is it possible to compute $P( F \mid S )$ given $P(F \mid S,A)$, $P(F \mid S, \lnot A)$?
This is a bit of a puzzle but you can compute a reasonable narrow limit even without knowing whether or not $P(S,A) = P(S) P(A)$.

Start with the contingency table relating $P(S, A)$, $P(S,\neg A)$, $P(\neg S, A)$, $P(\neg S,\neg A)$ to $P(S)$ and $P(A)$ :

$$\begin{array}{cc|c}
P( S,A)&amp; P(\neg S,A) &amp; P(A) \\
P(S,\neg A)&amp; P(\neg S,\neg A) &amp; P(\neg A) \\ \hline
P(S)&amp; P(\neg S) &amp; 1 \\
\end{array} \quad \rightarrow \quad \begin{array}{cc|c}
x&amp; 0.01-x &amp; 0.01 \\
0.07-x&amp; 0.92+x &amp; 0.99 \\ \hline
0.07&amp; 0.93 &amp; 1 \\
\end{array}$$

note that the cells must be between zero and one thus $0 \leq x \leq 0.01$



Then with

\begin{array}{rcrl}P(F|S) &amp;= &amp; &amp; P(F|S,A) P(A|S) + P(F| S,\neg A) P(\neg A|S)\\
&amp;=&amp;&amp; 1.0 \frac{x}{0.07} +  0.7 \frac{0.07-x}{0.07} \\
&amp; = &amp;&amp; 0.7 + 4 \frac{2}{7} x  \end{array}

you get

$$0.7 \leq P(F|S) \leq 0.743$$



To solve $P(F|S)$ exactly you need to narrow down $x$ more precisely. Possibilities are:  


If you know $P(F)$ then you could use $$\begin{array}{rcrl}P(F) &amp;= &amp; &amp; P(F|S,A) P(S,A) \\ &amp;&amp; +&amp; P(F|\neg S,A) P(\neg S,A) \\ &amp;&amp;+&amp; P(F| S,\neg A) P(S, \neg A)\\&amp;&amp;+&amp;P(F|\neg S,\neg A) P(\neg S,\neg A)\\
&amp;=&amp;&amp; 1.0 (x) + 0.7 (0.07 - x) + 0.9 (0.01-x)+0.1(0.92+x) \\
&amp; = &amp;&amp; 0.15-0.5 x \end{array}$$ 
If you know that $S \perp \! \! \! \perp A$ then you can use $x = P(S,A) = P(S)P(A) = 0.0007$